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\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+3H_2\\ Đặt:n_{Al}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}27a+56b=11\\1,5a+b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\ a,n_{HCl}=2.n_{H_2}=2.0,4=0,8\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,8}{8}=0,1\left(l\right)\\ b,FeCl_2+2AgNO_3\rightarrow Fe\left(NO_3\right)_2+2AgCl\downarrow\\ AlCl_3+3AgNO_3\rightarrow Al\left(NO_3\right)_3+3AgCl\downarrow\\ n_{AgCl}=n_{AgNO_3}=3.n_{AlCl_3}+2.n_{FeCl_2}=3.a+2.b=3.0,2+2.0,1=0,8\left(mol\right)\\ \Rightarrow a=\dfrac{170.0,8}{250}.100=54,4\%\\ b=m_{\downarrow}=m_{AgCl}=0,8.143,5=114,8\left(g\right)\)
\(a)Bte:3n_{Fe}+3n_{Al}=n_{Ag}\\ \Leftrightarrow n_{Ag_3}=0,1.3+0,1.3\\ \Leftrightarrow n_{Ag}=0,6mol\\ m_{rắn}=m_{Ag}=0,6.108=64,8g\\ BTNT\left(Ag\right):n_{Ag}=n_{AgNO_3}=0,6mol\\ V_{AgNO_3}=\dfrac{0,6}{2}=0,3l\\ BTNT\left(Al\right):n_{Al}=2n_{Al_2O_3}\\ \Leftrightarrow0,1=2n_{Al_2O_3}\\ \Leftrightarrow n_{Al_2O_3}=0,05mol\\ BTNT\left(Fe\right):n_{Fe}=2n_{Fe_2O_3}\\ \Leftrightarrow0,1=2n_{Fe_2O_3}\\ \Leftrightarrow n_{Fe_2O_3}=0,05mol \\ b=m_{oxit.bazo}=0,05.\left(160+102\right)=13,1g\)
Câu 1:
Đổi 500ml = 0,5l
nZn = 0,1 mol
Zn + 2HCl ---> ZnCl2 + H2
0,1 -----------------> 0,1 (mol)
=> VH2 =0,1*22,4 = 2,24l
Theo pt: nHCl = 2nZn = 0,2 mol
=> CM = n/V = 0,2/ 0,5 = 0,4M
2.
Ba(OH)2 + CO2 -> BaCO3 + H2O
nCO2=0,3(mol)
Theo PTHH ta có:
nCO2=nBaCO3=nBa(OH)2=0,3(mol)
mBaCO3=197.0,3=59,1(g)
CM dd Ba(OH)2=\(\dfrac{0,3}{0,6}=0,5M\)
\(n_{HCl}=\dfrac{3,65\%.100}{100\%.36,5}=0,1\left(mol\right)\)
Pt : \(2Na+2HCl\rightarrow2NaCl+H_2\)
0,15 0,1 0,1 0,05
Xét tỉ lệ : \(\dfrac{0,15}{2}>\dfrac{0,1}{2}\Rightarrow Nadư\)
\(m_{ddspu}=0,15.23+100-0,05.2=103,35\left(g\right)\)
\(C\%_{NaCl}=\dfrac{0,1.58,5}{103,35}.100\%=5,66\%\)
Chúc bạn học tốt
\(n_{HCl}=\dfrac{100.3,65}{100}:3,65=0,1mol\)
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
0,15 0,15 0,075
\(NaOH+HCl\rightarrow NaCl+H_2O\\ \Rightarrow\dfrac{0,15}{1}>\dfrac{0,1}{1}\Rightarrow NaOH.dư\\ n_{HCl}=n_{NaOH}=n_{NaCl}=0,1mol\\ m_{dd}=0,15.23+100-0,075.2=103,3g\\ C_{\%NaCl}=\dfrac{0,1.58,5}{103,3}\cdot100=5,66\%\\ C_{\%NaOH\left(dư\right)}=\dfrac{\left(0,15-0,1\right).40}{103,3}\cdot100=1,94\%\)
bài 1
\(n_{K_2CO_3}=\dfrac{100.13,8\%}{138.100\%}=0,1\left(mol\right)\)
\(n_{MgCl_2}=\dfrac{120.9,5\%}{95.100\%}=0,12\left(mol\right)\)
\(K_2CO_3+MgCl_2-->MgCO_3\downarrow+2KCl\)
\(\dfrac{0,1}{1}< \dfrac{0,12}{1}\Rightarrow\) K2CO3 hết MgCl2 dư
\(m_{MgCO_3}=0,1.84=8,4\left(g\right)\)
dd A :KCl và MgCl2 dư
\(m_{dd}=100+120-8,4=211,6\left(g\right)\)
\(C\%KCl=\dfrac{0,2.74,5}{211,6}.100\%\approx7,04\%\)
\(C\%MgCl_{2dư}=\dfrac{\left(0,12-0,1\right).95}{211,6}.100\%\approx0,9\%\)
bài 2
a) \(2X+nCl_2-->2XCl_n\)
a.......................................a
\(aX=6,72\left(1\right)\)
\(a\left(X+35,5n\right)=33,375\left(2\right)\)
\(\dfrac{\left(1\right)}{\left(2\right)}=\dfrac{aX}{a\left(X+35,5n\right)}=\dfrac{X}{X+35,5n}=\dfrac{6,72}{33,375}\)
\(\Rightarrow7,14n=0,78X\)
nếu n=1=>X=9,15(loại)
nếu n=2=>X=18,3(loại
nếu n=3=>X=27(chọn)
=> X là Al
b)
\(2Al+3H_2SO_4-->Al_2\left(SO_4\right)_3+3H_2\)
\(2Al+3CuSO_4-->Al_2\left(SO_4\right)_3+3Cu\)
\(2Al+2NaOH+2H_2O-->2NaAlO_2+3H_2\uparrow\)
Chất rắn k tan là Cu và có KL là 1g
\(Fe2O3+6HCl-->2FeCl3+3H2O\)
\(FeCl3+3NaOH-->Fe\left(OH\right)3+3NaCl\)
\(2Fe\left(OH\right)3-->Fe2O3+3H2O\)
\(n_{Fe2O3}=\frac{32}{160}=0,2\left(mol\right)\)
\(n_{Fe\left(OH\right)3}=2n_{Fe2O3}=0,4\left(mol\right)\)
\(n_{FeCl3}=n_{Fe\left(OH\right)3}=0,4\left(mol\right)\)
\(n_{Fe2O3}=\frac{1}{2}n_{FeCl3}=0,2\left(mol\right)\)
\(m_{Fe2O3}=0,2.160=32\left(g\right)\)
\(m=mFe2O3+m_{Cu}=32+1=33\left(g\right)\)
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
\(2FeCl_3+Cu\rightarrow CuCl_2+2FeCl_3\)
\(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\)
\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl\)
\(4Fe\left(OH\right)_2+2H_2O+O_2\rightarrow4Fe\left(OH\right)_2\)
\(Cu\left(OH\right)_2+CuO+H_2O\)
\(2Fe\left(OH\right)_3\rightarrow Fe_2O_3+3H_2O\)
\(a:Fe_2O_3\)
\(\Rightarrow n_{Cu_{pư}}=0,5a\left(mol\right)\)
\(160a+0,5a.80=32\)
\(\Rightarrow a=0,16\left(mol\right)\)
\(m_{Fe2O3}=0,16.160=25,6\)
\(m_{Cu}=0,08.64+1=6,12\left(g\right)\)
\(\Rightarrow m=25,6+6,12=31,72\left(g\right)\)
a/
\(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+2NaCl\)
\(Cu\left(ỌH\right)_2\rightarrow CuO+H_2O\)
b/
\(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\)
\(n_{Cu\left(ỌH\right)_2}=\dfrac{1}{2}n_{NaOH}=0,125\left(mol\right)\)
\(\rightarrow m_{Cu\left(OH\right)_2}=0,125.98=12,25\left(g\right)\)
\(n_{CuO}=n_{Cu\left(OH\right)_2}=0,125\left(mol\right)\)
\(\rightarrow m_{CuO}=0,125.80=10\left(g\right)\)
c/
\(V_{dd}=100+200=300\left(ml\right)=0,3\left(l\right)\)
\(n_{NaCl}=n_{NaOH}=0,25\left(mol\right)\)
\(\Rightarrow C_{M_{NaCl}}=\dfrac{0,25}{0,3}\approx0,83M\)
=> \(\dfrac{100}{200}=\dfrac{\left|10,4-C\right|}{\left|16,96-C\right|}\)
=> C% = 12,5867 %